To find the mass of copper(II) sulfate that remains after removing all the water, we need to first calculate the molar mass of CuSO4 · 5 H2O.
The molar mass of CuSO4 is 63.55 g/mol for copper, 32.07 g/mol for sulfur, and 4 * 16.00 g/mol for oxygen. Adding these together, we get a molar mass of 159.61 g/mol for CuSO4.
The molar mass of H2O is 2 * 1.01 g/mol for hydrogen and 16.00 g/mol for oxygen. Adding these together, we get a molar mass of 18.02 g/mol for H2O.
Next, we need to determine the molar mass of CuSO4 · 5 H2O by adding the molar mass of CuSO4 and 5 times the molar mass of H2O.
Molar mass of CuSO4 · 5 H2O = 159.61 g/mol + (5 * 18.02 g/mol) = 249.63 g/mol.
Now, we can calculate the number of moles of CuSO4 · 5 H2O in 89 g by dividing the given mass by the molar mass:
Number of moles = 89 g / 249.63 g/mol = 0.356 moles.
Since there are 5 moles of water in 1 mole of CuSO4 · 5 H2O, we multiply the number of moles by 5 to find the number of moles of water:
Number of moles of water = 0.356 moles * 5 = 1.78 moles.
To find the mass of water, we multiply the number of moles by the molar mass of water:
Mass of water = 1.78 moles * 18.02 g/mol = 32.06 g.
Finally, we subtract the mass of water from the given mass of CuSO4 · 5 H2O to find the mass of copper(II) sulfate remaining:
Mass of CuSO4 remaining = 89 g - 32.06 g = 56.94 g.
Therefore, the mass of copper(II) sulfate that would remain after removing all the water from 89 g of CuSO4 · 5 H2O is approximately 56.94 g.
